FSMQ
Advanced FSMQ
Additional Mathematics 6993
Mark Scheme for June 2010
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6993
Mark Scheme
June 2010
OCR - ADDITIONAL MATHEMATICS 6993
Marking instructions.
The total mark for the paper is 100.
Marks for method are indicated by an M. A method that is dependent on previous work is DM.
Marks for accuracy are of two kinds:
(i)
A mark indicates correct work only and
(ii)
F mark indicates that a "follow through" is allowed.
If an M mark is not gained then nor do any of the accuracy marks associated with it.
Marks not associated with a method are denoted B, which should be treated as "correct only", and
E which may be wrong because of a previous error.
Marks are not divisible except as indicated. e.g. A 2,1 means that 2 are awarded for a correct
answer and 1 for an answer that is only partially correct, as agreed at the meeting of Examiners.
When the method of solution is not one that has been discussed and does not fit the existing
scheme then an alternative scheme should be devised which maintains the same number of M, A,
F, B and E marks. You should also bring this to the attention of the Principal Examiner.
The rubric says that the norm is for answers to be given to 3 s.f. except where indicated. Where
this rubric is ignored then 1 mark should be deducted once in the paper, at the point where it is
first met. This should be indicated -1, TMSF or -1TFSF. Details will be discussed at the meeting
of examiners.
Misreading of a question (including the candidate's own working) should normally be penalised
by the loss of the relevant accuracy mark or two marks (whichever is less); but if the question is
made substantially easier then further penalties may be imposed.
Sub-marks should be shown near to the relevant work. If these are individual marks then the
appropriate letter should be given. Sub-marks are given in the question paper and the mark
scheme. For substantially correct solutions a number of sub-marks may be combined, even up to
the total mark for the question for a totally correct question. The sum of the sub-marks are then
added and ringed at the end of the question. (This means that a totally correct question has the
total mark written twice - once as a "sum of sub-marks" and unringed and once ringed as the total
for the question.) The total mark for the paper should be given on the front page, top right and
ringed.
Work that is crossed out and not replaced should be marked. If work has been crossed out and
replaced then the replacement work should be marked even if it is incorrect and the crossed out
work correct.
Any notation that is understandable may be used to support your marking. In particular:
isw – ignore subsequent working
www – without wrong working
soi – seen or implied
1
6993
Mark Scheme
June 2010
An independent person should be used to check the summation of marks. You should add the
ringed marks on the paper to check the addition and the independent checker should add the
unringed marks. There is a fee paid for this checking - if you are unable to find anyone to do this
work the Board and the Principal Examiner must be informed.
Please mark in red.
If examiners have any doubt about the interpretation of any instructions or if any point of
difficulty arises during the marking of scripts, they should communicate with the Principal
Examiner.
2
6993
Mark Scheme
June 2010
Section A
3 − x < 4( x − 1)
 3 − x < 4x − 4
 7 < 5x
7
x>
5
1
B1
B1
Sight of 4x – 4
Sight of ax and b where
either a = 5 or b= 7 oe
B1
Final answer WWW
3
B1
B1
12 
12 
12 
= 1 −   x +   x 2 −   x3
1
2
3
2
3
= 1 − 12 x + 66 x − 220 x
2
B1
Ignore terms of higher power
3
(i)
3
M1
A1
Remainder is f(−1)
= −1 − 5 − 2 + 8 = 0
For long division x3 + x2 in working and x2 in
quotient must be seen for M1
Or by inspection (x + 1)(x2 + …) for M1
(ii)
Signs and powers
2 out of 3 coefficients
worked out
All coefficients and 1
Or long division
0 must be seen or
implied
2
x3 − 5 x 2 + 2 x + 8 = 0
 ( x + 1) ( x 2 − 6 x + 8 ) = 0
M1
 ( x + 1)( x − 2 )( x − 4 ) = 0
Factorise cubic to give
(x + 1)(ax2 + bx + c)
Solve their quadratic
DM1
 x = −1, 2, 4
A1
Allow ans with no working
3
Alt: Trial to find one root: x = 2, 4 M1, A1
 x = –1, 2, 4
A1
3
6993
4
(i)
Mark Scheme
June 2010
M1
4
625
5
= 0.4823
  =
 6  1296
Either form or 0.482
isw
A1
2
(ii)
4
3
5
5 1
1−   − 4   
6
6 6
625 500
= 1−
−
= 1 − 0.4823 − 0.3858
1296 1296
171
19
=
=
= 0.1319
1296 144
A1 ans
4
1 − 2 terms
4 soi
Powers
A1
Ans in either form or
0.132
4
M1
B1 both
coeffs
B1 powers
Alt: Add three terms
2
2
3
4
5 1
 5  1   1 
6    + 4   +  
6 6
 6  6   6 
= 0.11574 + 0.01543 + 0.00077
= 0.1319
M1
B1
B1
6993
5
(i)
Mark Scheme
June 2010
M1
Diffn and set = 0
A1
Derived fn
When x = −1, y = 12
A1
Stationary point
d2 y
= 6 x − 6 < 0 when x = −1 so maximum
dx 2
M1
A1
To find nature of
turning points
dy
= 3x 2 − 6 x − 9
dx
= 0 when 3x 2 − 6 x − 9 = 0  x 2 − 2 x − 3 = 0
 ( x − 3)( x + 1) = 0  x = 3, −1
Allow SC1 for (–1, 12) with no working
Alternative ways to demonstrate maximum at
x = –1
Value of y
–1–
y < 12
–1
y = 12
–1 +
y < 12
–1
dy
=0
dx
(i)
–1 +
dy
<0
dx
A1
u = 90, v = 6, s = 2016
 62 = 902 + 2a × 2016
a=−
Allow at most one
integer either side
(typically, x = –2, 0 if
turning point is correct)
M1
(ii)
6
M1
A1
Gradient of tangent
–1–
dy
>0
dx
5
902 − 62
8064
=−
= −2 m s −2
4032
4032
B1
General shape:
turning points in correct
quadrants
1 Intercept on y axis in
[0,12]
Does not turn back on
itself.
M1
A1
A1
Using correct formula
Correct substitution
3
(ii)
u = 90, v = 6, a = −2
 6 = 90 − 2t
84
t =
= 42 secs
2
The two parts can be the other way round
5
M1
Using correct formula
A1
2
6993
7
(i)
Mark Scheme
sin θ cos θ sin 2 θ + cos 2 θ
+
=
cos θ sin θ
sin θ cos θ
1
=
sin θ cos θ
June 2010
B1
1
Alt:
sin 2 θ + cos 2 θ = 1
cos 2 θ
1
=
sin θ sin θ
sin θ cos θ
1

+
=
cos θ sin θ sin θ cos θ
1
sin θ cos θ
sin θ cos θ = 
+
=4
4
cos θ sin θ
1
 tan θ +
=4
tan θ
 sin θ +
(ii)
(iii)
M1
A1
2
1
tan θ +
= 4  tan 2 θ + 1 = 4 tan θ
tan θ
2
 t − 4t + 1 = 0
M1
Clear fractions to give 3
term quadratic
4 ± 16 − 4
= 2 ± 3 ( = 3.732 and 0.268 )
2
 θ = 150 and 750
M1
Sub numbers into
correct quadratic
3sf or more
Rounds to these
t=
8
Using (i) and tan
A1
A1
Sp Case B1 for 15 and B1 for 75 with no supporting
working
v = 60 ( t 4 − 10t 3 + 25t 2 )
4
5
 s =  ( 60t 4 − 600t 3 + 1500t 2 ) dt
0
= 12t 5 − 150t 4 + 500t 3 
5
0
= 6250 m
If 60 is left out then 4/5 only.
M1
A2,1
Integrate
Terms 1 each error
DM1
A1
Sub t = 5
Cao
5
6
6993
9
Mark Scheme
(i)
June 2010
For 8 WWW
For 2 WWW
B1
B1
 1 + 15 3 + 1 
Centre is 
,
 = (8, 2)
2 
 2
2
Nb Working with vectors to give diameter = [14,2]
and so radius = [7,1] giving centre (15 – 7, 3 –1) is
correct.
(ii)
M1
A1
PC = (8 − 1) 2 + (2 − 3) 2 = 50 = 5 2
For
50
2
Alt:
Length of diameter =
(15 − 1) + ( 3 − 1)
2
2
= 142 + 22
= 200 = 10 2
 Radius = 5 2
(iii)
( x − 8) + ( y − 2 )
2
2
M1
= 50
Correct use of formula
including 50 and using
their midpoint.
 x 2 + y 2 − 16 x − 4 y + 64 + 4 − 50 = 0
 x 2 + y 2 − 16 x − 4 y + 18 = 0
A1
2
10
(i)
(ii)
(iii)
Sub (0,4)
M1
1
Gives k =
2
A1
Sub (0, 4)
M1
1
Gives c = −
4
A1
2
2
1
When x = 3 y = − (3 − 2)2 (3 − 4) = 0.25 for cubic
4
Or when x= 3, y > 0 for cubic
B1
John’s model is better
DB1
2
7
6993
Mark Scheme
June 2010
Section B
Allow 4 sf in this question
11 (i)
M1
A1
AF
BF
100
=
=
sin70 sin60 sin 50
100
 AF =
× sin 70 ( = 122.7 m )
sin 50
100
 BF =
× sin 60 = 113.1 m oe
sin 50
Sin rule applied
Sight of 50 and 70
A1
M1
A1
Correct sine rule to find
BF
5
(ii)
Alt:
Cosine rule for BF:
BF2 = 1002 + 122.72 – 2×100×122.7×cos60
= 12785
BF = 113.1
FT = AF × tan10
= 122.7 tan10 = 21.6 m
M1
A1
M1
A1
2
Anything that rounds to 21.6
(iii)
CF = 122.7sin60
=106.3 m
Or: = their BF ×sin70
M1
A1
TheirFT
TheirCF
 θ = 11.50
M1
F1
 tan θ =
Accept 106.2 or 106
Using tan correctly
Substituting correctly
Accept 11 or 12
A1
5
Alt: to find CF.
Area of triangle = 12 × AF . AB sin 60 = 5313
M1
 × CF ×100 = 5313  CF = 106.3
A1
1
2
8
6993
12
Mark Scheme
(i)
June 2010
y = 0.3 x 2 − 1.5 x
dy
= 0.6 x − 1.5
dx
When x = 5 gt = 1.5
2
3
2
AB: y = − ( x − 5 )
3
 2 x + 3 y = 10
 gn = −
B1
Derivative
M1
Find g t and use of m 1 ×
m 2 = –1
For g n
A1
Line in any simplified
form
A1
4
(ii)
Solve simultaneously:
3 y + 2 x = 10
M1
Method to eliminate
one variable
2 y + 3x = 0
6 y + 4 x = 20
6 y + 9x = 0
5 x = −20
F1
A1
 x = −4, y = 6
SC1: answer with no working
(iii)
Area of triangle =
3
1
× 5 × their y = 15
2
5
Area under curve =
 ( 0.3x
0
2
x and y.
− 1.5 x ) dx
=  0.1x 3 − 0.75 x 2 
= −6.25
 Area of card = 15 + 6.25 = 21.25
Other methods, follow scheme
ieE1 Area of triangle
M1 area as integral
A1 Integrand
A1 value for area
A1 Final answer
9
5
E1
Might appear anywhere
in this part
M1
Ignore limits here
A1
0
A1
Condone lack of –ve
sign
A1
5
6993
13 (i)
(ii)
Mark Scheme
Ali:
72
t
Beth:
June 2010
M1
72
A1 A1 Accept Beth: t − 3
3
M1
Subtraction of their
terms = 3
A1
M1
Multiply out
A1
A1
and simplify
5
72
t+2
72 72
−
=3
t t+2
 72(t + 2) − 72t = 3t (t + 2)
 72t + 144 − 72t = 3t (t + 2)
 3t (t + 2) = 144
Alternative (based on alternative answer to (i)
72
=t+2
72
−3
t
 72t = ( 72 − 3t ) (t + 2)
M1 A1
 72t = 72t − 3t 2 + 144 − 6t
(iii)
M1
 3t 2 + 6t = 144  3t (t + 2) = 144
3t (t + 2) = 144
A1
A1
 3t 2 + 6t − 144 = 0
M1
 t 2 + 2t − 48 = 0
A1
 (t + 8)(t − 6) = 0
t =6
A1
 Ali takes 6 hours and Beth takes 8 hours.
A1
For quadratic in
simplified form. (See
below)
www
4
SC1 for answer with no working
What is “simplified form”?
Either a quadratic with all three terms on left = 0 ready for the use of the
formula
OR:
Divide through by 3 giving t2 + 2t = 48 ready for solving by the
completion of the square.
10
6993
14
Mark Scheme
(i)
200x + 100y ≥ 1500 oe
June 2010
M1
A1
Deriving a linear
inequality
2
(ii)
y≥x
B1
1
(iii)
One line
Other line
Shading for both, ft
their inequalities
B1
B1
E1
No Scales: B0, B0, E1
Condone scales not as
instructed.
3
(iv) C = 80x + 60y
B1
Correct point is (5, 5)
Cost = £700
In absence of OF, 80 ×5 + 60× 5 must be seen
B1
M1
A1
Now minimum cost is at (7, 1)
Giving £620
Nb (8, 0) gives £640
B1
B1
Sub in OF
4
(v)
2
11
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